Congruum Problem

Date: 04/04/2002 at 18:06:35

From: Dr. Allan

Subject: Congruum problem

In researching suitable problems for my students I have found a

reference to Fibonacci and his congruum problem. Something has me

stumped; I hope one or more of you can tell me what I am doing wrong:

Quote from the article at the MacTutor History of Mathematics archive

at St. Andrews University:

http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Fibonacci.html

"[Fibonacci] defined the concept of a congruum, a number of the form

ab(a + b)(a - b), if a + b is even, and 4 times this if a + b is odd

where a and b are integers. Fibonacci proved that a congruum must be

divisible by 24 and he also showed that for x,c such that x^2 + c and

x^2 - c are both squares, then c is a congruum. He also proved that a

square cannot be a congruum."

With x=15 and c=216, we get the two squares 441 and 9, meaning that

216 should be a congruum. Thus we should be able to find numbers a

and b such that

216 = ab(a+b)(a-b) if a+b is even

or

54 = ab(a+b)(a-b) if a+b is odd

Now, b must be smaller than a and d:= ab(a+b)(a-b) is increasing in

both a and b.

Let's look at the possible scenarios.

Say a+b is odd. We want d to equal 54. Our triples (b,a,d) give us

(1,2,6) (1,4,60); thus b=1 is not possible

(2,3,30) (2,5,210); thus b=2 is not possible

(3,4,84); thus b = 3 and b > 3 is not possible (since d is

increasing); thus a+b odd is not possible

Say a+b is even. We want d to equal 216. Our triples (b,a,d) give us

(1,5,120) (1,7,336); thus b=1 is not possible

(2,4,96) (2,6,3849; thus b=2 is not possible

(3,5,360); thus b = 3 and b > 3 is not possible (since d is

increasing); thus a+b even is not possible

So, because 15^2 + 216 = 21^2 and 15^2 - 216 = 3^2 we should have

that 216 is a congruum, but we cannot put it in the form ab(a+b)(a-b).

What is wrong with this argument?

Sincerely,

Dr. Allan

--------------------------------------...

Date: 04/05/2002 at 16:04:44

From: Dr. Paul

Subject: Congruum problem

I went to the library this morning and picked up a copy of Ore's

_Number Theory and its History_. The solution to the congruum problem

is listed on pages 188-193 and I found it very interesting reading.

What follows is essentially Ore's solution with my commentary added

where I felt Ore's details were (probably intentionally) lacking

sufficient explanation:

We want to find a number x such that simultaneously:

x^2 + h = a^2, x^2 - h = b^2 (equation 8-16)

and determine for which h rational solutions x can exist. We shall

first determine the solutions in integers, and this depends, as we

shall see again, on the Pythagorean triangle. When the second equation

in (8-16) above is subtracted from the first, one has:

2*h = (a^2 - b^2) = (a-b)*(a+b) (8-17)

Since the left-hand side is even, a and b must both be odd or both be

even. Therefore, a - b is even. Hence

a - b = 2*k

and k must be a divisor of h since according to (8-17) we have:

2*h = 2*k*(a+b)

or

h = k*(a+b)

which implies k divides h.

It follows that

(a + b) = h/k

and by adding and subtracting the last two equations, one finds

a = h/(2*k) + k, b = h/(2*k) - k

When these two expressions are substituted into the original equations

(8-16), we obtain:

x^2 + h = [h/(2*k) + k]^2 = [h/(2*k)]^2 + h + k^2

and

x^2 - h = [h/(2*k) - k]^2 = [h/(2*k)]^2 - h + k^2

Adding these equations and removing a common factor of two gives:

x^2 = [h/(2*k)]^2 + k^2

In our case of interest, we have:

x = 15

h = 216

a = 21

b = 3

k = 9

Therefore, the three numbers

x, h/(2*k), k

form a Pythagorean triangle. Determine the lesser of h/(2*k) and k (in

this case k = 9 < 12 = 216/18 = h/(2*k)) so that we can write:

x = t*(m^2 + n^2)

k = t*(m^2 - n^2) (**)

h/(2*k) = 2*m*n*t

where t is some integer and the expressions in m and n define a

primitive solution of the triangle.

In actuality, we have:

x = 15

k = 9

h = 216

so clearly, t = 3 and this forces m = 2, n = 1

As an aside, if h/(2*k) > k then we pick

h/(2*k) = t*(m^2 - n^2)

k = 2*m*n*t

Now we take the product of the last two expressions in (**) above to

obtain as the general solution to (8-16)

x = t*(m^2 + n^2), h = 4*m*n*(m^2 - n^2)*t^2 (8-18)

Now we make a slight reduction to this solution. Let us suppose that

we have a solution x of (8-16), where x has the factor t and h at the

same time has the factor t^2 (this is the case in our example since

x = 15 is divisible by three and h = 216 is divisible by 9). Then we

can write:

x = x_1 * t, h = h_1 * t^2

From (8-16), we obtain:

[(x_1)^2 * t^2] + (h_1 * t^2) = a^2

and

[(x_1)^2 * t^2] - (h_1 * t^2) = b^2

Then a^2 and b^2 are both divisible by t^2 so a and b are both

divisible by t:

a = a_1 * t, b = b_1 * t

This gives:

[(x_1)^2 * t^2] + (h_1 * t^2) = [(a_1)^2 * t^2]

and

[(x_1)^2 * t^2] - (h_1 * t^2) = [(b_1)^2 * t^2].

Cancelling the factor of t^2 gives:

(x_1)^2 + h_1 = (a_1)^2

and

(x_1)^2 - h_1 = (b_1)^2

When no further reduction is possible, we shall say that we have a

primitive solution. When this reduction is applied to (8-18), we

obtain:

x = m^2 + n^2

h = 4*m*n*(m^2 - n^2)

The numbers m and n now produce a primitive Pythagorean triple.

Thus your solution of

15^2 + 216 = 21^2

15^2 - 216 = 3^2

Is really just a multiple of the primitive solution:

5^2 + 24 = 7^2

5^2 - 24 = 1^2

This corresponds to the first row in the table here:

Congruum Problem - MathWorld - Eric Weisstein

http://mathworld.wolfram.com/CongruumProblem.html

i.e., take m = 2, n = 1. Then x = 5, h = 24, a = 7, b = 1

and as with Pythagorean triples, letting t be a natural number gives

an infinite number of solutions.

Neat problem!

I hope this is clear - let me know if I'm skipping steps where I need

to be filling in the details.

Finally, notice that writing

h = 4*m*n*(m-n)*(m+n)

is only possible when h is an element of a primitive congruum. That's

why your attempts failed. You needed to write

h = 4*m*n*(m-n)*(m+n)*t^2

which works easily when h = 216, m = 2, n = 1, t = 3.

- Doctor Paul, The Math Forum

http://mathforum.org/dr.math/

Source(s):
http://mathforum.org/library/drmath/view/59276.html